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6Coupled OscillatorsIn what follows, I will assume you are familiar with the simple harmonic oscilla-

tor and, in particular, the complex exponential method for finding solutions of the oscillator equation of motion. If necessary, consult the revision section on Simple

Harmonic Motion in chapter 5.

6.1 Time Translation Invariance

Before looking at coupled oscillators, I want to remind you how time translation invariance leads us to use (complex) exponential time dependence in our trial solu- tions. Later, we willsee that spatial translationinvarianceleads to exponentialforms for the spatial parts of our solutionsas well. To examine theimplicationoftimetranslationinvariance,it'senoughtoconsider a single damped harmonic oscillator, with equation of motion, m¨x ??2mgx?mw20x? where the two terms on the right are the damping and restoringforces respectively.

We can rearrange this to,

¨x ?2gx?w20x?0? To solve this equation, we used an ansatz (or guess) of the form x ?AeWt? whereAandWare in general complex (to get a physical solution you can usethe real or imaginary parts of a complex solution). The reason that we could guess such a solution lies in time translation invariance. What this invariance means is that we don't care about the origin of time. It doesn't matter what our clock read when we started observingthe system. In the differential equation, this property appears because the time dependence enters only through time derivatives,notthrough the value of time itself. In terms of a solution x ?t?, this means that: ifx?t?is a solution,then so isx?t?c?for any constantc. Thesimplest possibilityis thatx?t?c?is proportionaltox?t?, withsome proportion- ality constantf ?c?, depending onc, x ?t?c? ?f?c?x?t?? 55

566 Coupled Oscillators

We can solve this equation by a simple trick. We differentiate with respect tocand then setc ?0 to obtain x ?t? ?Wx?t?? whereWis just the value off?0?. The general solution of this linear first order differential equation is x ?t? ?AeWt? We often talk aboutcomplexexponential forms becauseWmust have a non-zero imaginary part if we want to get oscillatory solutions. In fact, from now on I will let W ?iw, so thatwis real for a purely oscillatory solution. We can't just useanyvalue we like forw. The allowed values are determined by demanding thatAeiwtactually solves the equation of motion: ??w2?2igw?w20?Aeiwt?0? Ifwearetohaveanon-trivialsolution,Ashouldnotvanish. Thefactorinparentheses must then vanish, giving a quadratic equation to determinew. The two roots of the quadratic give us two independent solutionsof the originalsecond order differential equation.

6.2 Normal Modes

We want to generalise from a singleoscillatorto a set of oscillatorswhich can affect each others' motion. That is to say, the oscillators arecoupled.

If there arenoscillators with positionsxi

?t?fori?1? ? ? ? ?n, we will denote the "position"of the whole system by a vectorx ?t?of the individuallocations: x ?t? ? ?x1?t??x2?t?? ? ? ? ?xn?t??? The individualpositionsxi?t?might well be generalised coordinates rather than real physical positions. The differential equations satisfied by thexiwill involve time dependence only through time derivatives, which means we can look for a time translation invariant solution, as described above. This means all the oscillators must have the same complex exponential time dependence,eiwt, wherewis real for a purely oscillatory motion. The solution then takes the form, x ?t? ? 0 B B? A1 A 2... A n 1C CA eiwt? where theAiareconstants. This describes a situation where all the oscillators move with thesame frequency, but, in general, different phases and amplitudes: the oscil- lators' displacements are in fixed ratios determined by theAi. This kind of motion is called anormal mode. Theoverallnormalisation is arbitrary (by linearity of the differential equation), which is to say that you can multiply all theAiby the same constant and still have the same normal mode. Our job is to discover whichware allowed, and then determine the set ofAi corresponding to each allowedw. We will find precisely the right number of normal modes to provide all the independent solutions of the set of differential equations. Fornoscillators obeying second order coupled equations there are 2nindependent

6.3 Coupled Oscillators57

solutions: we will findncoupled normal modes which will give us 2nreal solutions when we take the real and imaginary parts. Once wehave found all thenormal modes, we can constructanypossiblemotion of the system as a linear combination of the normal modes. Compare this with Fourier analysis, where any periodic function can be expanded as a series of sines and cosines.

6.3 Coupled Oscillators

Take a set of coupled oscillators described by a set of coordinatesq1 ? ? ? ? ?qn. In general the potentialV ?q?will be a complicated function which couples all of these oscillators together. Considersmalloscillations about a position of stable equilib- rium, which (by redefining our coordinates if necessary) we can take to occur when q i ?0 fori?1? ? ? ? ?n. Expanding the potential in a Taylor series about this point, we find, V ?q? ?V?0? ?å 0q i ?1 2å i 0q iqj By adding an overall constant toVwe can chooseV?0? ?0. Since we are at a position of equilibrium, all the first derivative terms vanish. So the first terms that contributeare the second derivative ones. We define, K ij 0 and drop all the remaining terms in the expansion. Note thatKijis a constant sym- metric (why?)n ?nmatrix. The corresponding force is thus F i jK ijqj and thus the equations of motion are M i¨qi jK ijqj fori?1? ? ? ? ?n. Here theMiare the masses of the oscillators, andKis a matrix of `spring constants'. Indeed for a system of masses connectedby springs, with each mass moving in the same single dimension, the coordinates can be taken as the real positioncoordinates, and thenMis a (diagonal in this case) matrix of masses, while Kis a matrix determined by the spring constants. Be aware however, that coupled oscillator equations occur more generally (for example in electrical circuits) where theqis need not be actual coordinates but more general parametersdescribing the system (known as generalised coordinates) and in this caseMandKplay similar rˆoles even if they do not in actuality correspond to masses and spring constants. To simplify the notation,we will writethe equations of motion as a matrix equa- tion. So we define, M 0 B B?

M10? ? ?0

0M2 ? ? ?0............ 0 0 ? ? ?Mn 1C CA K? 0 B B?

K11K12? ? ?K1n

K 21K22
? ? ?K2n............ K n1Kn2 ? ? ?Knn 1C CA

586 Coupled Oscillators

x1x2 m

1m2k1k

0k2 Figure 6.1Two coupled harmonic oscillators. The vertical dashed lines mark the equilib- rium positions of the two masses.

Likewise,letqand¨qbe column vectors,

q 0 B B? q1 q 2... q n 1C CA ¨q 0 B B?

¨q1

¨q2...

¨qn

1C CA

With this notation, the equation of motion is,

M ¨q ??Kq?or¨q??M ?1Kq whereM ?1is the inverse ofM.

Now look for a normal mode solution,q

?Aeiwt, whereAis a column vector.

We have

¨q ??w2q, and cancellingeiwtfactors, gives finally, M ?1KA ?w2A? This is now aneigenvalue equation. The squares of the normal mode freqencies are theeigenvaluesofM ?1K, with thecolumn vectorsAas the correspondingeigenvec- tors.

6.4 Example: Masses and Springs

As a simple example, let's look at the system shown in figure 6.1, comprising two massesm1andm2constrained to move along a straight line. The masses are joined by a spring with force constantk

0, andm1(m2) is joined to a fixed wall by a spring

with force constantk1(k2). Assume that the equilibrium position of the system has each springunstretched,and use thedisplacementsx1andx2ofthe twomasses away from their equilibrium positionsas coordinates. The forceon massm1is then F 1 ??k1x1?k

0?x1?x2?

and on massm2 F 2 ??k2x2?k

0?x2?x1??

(Note that these followfrom a potential of formV?1

2k1x21

?1 2k

0?x2?x1?2

?1

2k2x22.)

You can check that Newton's 2nd law thus implies, in matrix form: ?m10quotesdbs_dbs20.pdfusesText_26